1001 Algebra Problems.PDF

558. b.To begin, we must make certain that the
     numerator and denominator are both com-
     pletely factored and that all common terms are
     canceled:

[xx+–3^5 – x]x–^1  3 = [xx+–3^5 – x(xx–– 3 3)]x–^1  3 =

(x– 3) = – (x^2 – 4x– 5) =

–(x–5)(x+ 1)

Determine the x-values that make this expres-

sion equal to zero or undefined.Then, we assess

the sign of the expression on the left side on

each subinterval formed using these values. To

this end, observe that these values are x= –1,

3, 5. Now, we form a number line, choose a

real number in each subinterval, and record

the sign of the expression above each:

Since the inequality does not include “equals,”

we would not include those values from the

number line that make the numerator equal to

zero. As such, we conclude that the solution set

is [–1, 3)∪(3, 5).

559. d.Our first step is to make certain that the
     numerator and denominator are both com-
     pletely factored and that all common terms
     are canceled:

 2 xx+1–  2 x^1 –1+ =  –

 + = =

= =

Now, the strategy is to determine the x-values

that make this expression equal to zero or

undefined. Then, we assess the sign of the

expression on the left side on each subinterval

formed using these values. To this end, observe

that these values are x= –^12 ,–^14 ,^12 , and 1. Next,

form a number line, choose a real number in

each subinterval, and record the sign of the

expression above each:

Since the inequality includes “equals,” we

include those values from the number line that

make the numerator equal to zero. The solution

set is (–^12 ,–^14 ]∪(^12 , 1].

560. d.First, we must make certain that the numer-
     ator and denominator are both completely fac-
     tored and that all common terms are canceled,
     as follows:

(^3 yy–+ 12 ) 2 – (y–^71 y)(–y^3 +1)+ = 

yy++1 1 – (y–^71 y)(–y^3 +1)yy––1 1 + y+^5  1  =

=

=

=

= =

Now, the strategy is to determine the y-values

that make this expression equal to zero or

undefined. Then, we assess the sign of the

expression on the left side on each subinterval

formed using these values. To this end, observe

that these values are y = –4, –1, 1. Next, we

form a number line, choose a real number in

y+ 4

(y– 1)^2

(y+4)(y+ 1)

(y– 1)^2 (y+ 1)

y^2 + 5y+ 4

(y– 1)^2 (y+ 1)

^3 y^2 + 5y+ 2 – 7y^2 + 10y– 3 + 5y^2 – 10y+ 5

(y– 1)^2 (y+ 1)

^3 y^2 + 5y+ 2 – (7y^2 – 10y+ 3) + 5(y^2 – 2y+ 1)

(y– 1)^2 (y+ 1)

(3y+ 2)(y+ 1) – (7y–3)(y– 1) + 5(y–1)^2

(y– 1)^2 (y+1)

(y–1)^2

(y– 1)^2

^3 y+ 2

(y– 1)^2

^5

(y+ 1)

1

+ – + +–

-^121412

(4x+ 1)(x– 1)

(2x– 1)(2x+ 1)

^4 x^2 – 3x– 1

(2x– 1)(2x+ 1)

^2 x^2 – x– 2x– 1 + 2x^2

(2x– 1)(2x+ 1)

x(2x– 1) – 1(2x+ 1) + 2x^2

(2x– 1)(2x+ 1)

^2 x^2

4 x^2 – 1

^2 x+ 1

2 x+ 1

^1

2 x– 1

^2 x– 1

2 x– 1

x

2 x+ 1

^2 x^2

4 x^2 – 1

1 3 5

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x+ 5 – x^2 + 3x

x– 3

ANSWERS & EXPLANATIONS–